# Question Paper

Pages: 14 (4506 words) Published: June 1, 2013
PART- A : PHYSICS
1. A uniform cylinder of length L and mass M having cross - sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at equilibrium position. The extension x0 of the spring when it is in equilibrium is : (1) Mg k (2) Mg LA 1 kM σ ⎛⎞ − ⎜⎟ ⎝⎠ (3) Mg LA 1 k 2M σ ⎛⎞ − ⎜⎟ ⎝⎠ (4) Mg LA 1 kM σ ⎛⎞ + ⎜⎟ ⎝⎠ (Here, k is spring constant) 1. (3) 0 LA Kx g mg 2 ⎛⎞ + σ = ⎜⎟ ⎝⎠ ∴ x0 = Mg LA 1 k 2M α ⎛⎞ − ⎜⎟ ⎝⎠ 2. A metallic rod of length ‘’ is tied to a string of length 2 and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is : (1)

22B 2 ω (2) 2 3B 2 ω (3) 2 4B 2 ω (4) 2 5B 2 ω
2. (4) V = () 3 2 v B d x Bdx × ⋅ = ω ∫∫
V =
25B 2 ω
3. This question has statement I and Statement II. Of the four choice given after the statements, choose the one that best describes the two statements. Statement – I : A Point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as 21m f mv thenf 2 M m ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ + ⎝ ⎠ ⎝ ⎠ . Statement – II : Maximum energy loss occurs when the particles get stuck together as a result of the collision. (1) Statement - I is true, Statement - II is true, statement - II is a correct explanation of Statement - I (2) Statement - I is true, Statement - II is true, statement - II is not a correct explanation of Statement – I (3) Statement - I is true, Statement - II is false (4) Statement – I is false, Statement – II is true 2

x
V
ω
ga
JEE-MAIN 2013 : Paper and Solution (4)
(Pg. 4)
3. (4) Maximum energy loss when inelastic collision takes place mv = (m + M) v′ v′ = m v mM + 2 i 1 k mv 2 =
22 22 f 2 1 1 m v 1 m k (m M)v' (m M) mv 2 2 2 M m (m M) ⎛⎞= + = + = ⎜⎟ ++ ⎝⎠ Loss of energy = 22 if 1 m M 1 k k mv 1 mv 2 M m M m 2 ⎡⎤ − = − = × ⎢⎥ ++ ⎣⎦ 4. Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = Time and A = electric current, then : (1) [ε0] = [M−1 L−3 T2 A] (2) [ ε0] = [M−1 L−3 T4 A2] (3) [ε0] = [M−1 L2 T−1 A−2] (4) [ ε0] = [M−1 L2 T−1 A] 4. (2) [] 2 2 2 0 22 2 C A T Nm MLT L − ⎡⎤ ε = = ⎢⎥ − ⎣⎦ [ ] 1 3 4 2 0 M L T A −−ε= 5. A Projectile is given an initial velocity of ˆˆ (i 2j)m/s, + where ˆ i is along the ground and ˆ j is along the vertical. If g = 10 m/s2, the equation of its trajectory is : (1) y = x − 5x2 (2) y = 2x − 5x2 (3) 4y = 2x − 5x2 (4) 4y = 2x − 25x2 5. (2) u = 22 2 1 5 + = θ = tan−1 2 Equation : y = x tan θ − 2 22 gx 2u cos θ y = 2x −

210x
1
2
5
× 5 ×
y = 2x − 5x2
6. The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to α times its original magnitude, where α equals : (1) 0.7 (2) 0.81 (3) 0.729 (4) 0.6 6. (3) A = Ao e−kt 0.9Ao = Aoe−kt −kt = n (0.9) −5 k = n (0.9) ⇒ −15 k = 3 n (0.9) A = Aoe−15k = Ao 3 n(0.9)e − = (0.9)3 Ao = 0.729 Ao (5) VIDYALANKAR : JEE-MAIN 2013 : Paper and Solution

(Pg. 5)
7. Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then : (1) 5C1 = 3C2 (2) 3C 1 = 5C2 (3) 3C 1 + 5C2 = 0 (4) 9C1 = 4C2 7. (2) Potential = 0 on connecting them together i.e. Q = 0 i.e. C1V1 = C2V2 [capacitance is positive but they are connected with opposite polarity] 120 C1 = 200 C2 3C1 = 5C2 8. A sonometer wire of length 1.5 m is made of steel. The tension in it produce an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg / m3 and 2.2 × 1011 N/m2 respectively ? (1) 188.5 Hz (2) 178.2 Hz (3) 200.5 Hz...